🚀 go-pugleaf

QUESTION: Suppose you have an n-element multiset S deforming into another
n-element multiset T.  Can you find a path from a given element of S to some
element of T?

More precisely, let M be a topological space and give M^n the product topology.
Let the symmetric group S_n act on M^n by permuting the coordinates, and give
M^n/S_n the quotient topology.  (M^n/S_n is the space of n-element multisets.)
Suppose w : [0,1] -> M^n/S_n is a continous path, and P is an element of the
multiset w(0).  Can one find a path w_1 : [0,1] -> M such that w_1(0)=P and
w_1(t) is an element of w(t) for all t in [0,1]?

Assume that M has nice properties, if it helps.  I need the case M=R^2 (i.e.,
the plane).  (The application is the following: take all polynomials
	1 + e_1 z + e_2 z^2 + ... + e_n z^n
where each e_i is 0 or 1, and n is arbitrary.  Let W be the set of all complex
zeros of such polynomials.  I have proved that the closure of W is connected.
If the question above has a positive answer, I can prove that it is path
connected as well.)

More generally one can ask whether the path w : [0,1] -> M^n/S_n can always be
decomposed into n paths w_1,w_2,...,w_n in M.  This is equivalent to asking
whether the path can be lifted to a path in M^n.  (This would certainly imply a
positive answer to the first question.)

Here is what I know so far.  If M=R this general result is true.  (Let w_k(t)
be the k^th largest element of w(t).)  If M=R^2 and n=2, the result is true.
(The map M^n -> M^n/S_n is a covering projection if one deletes the places
where two elements of the multiset coincide, so one can lift the path on each
(open) connected component of the set of t for which w(t) has distinct
elements.  For n=2, this suffices, since for the remaining values of t, there
is only one possible lift of w(t).)

Please e-mail solutions, useful references, or comments to me at

	poonen@research.att.com		before August 24
	poonen@math.berkeley.edu	after August 24

Bjorn Poonen